It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm

Perimeter = 5 + 5 + 1

= 11 cm

So, semi perimeter = 11/2 cm

= 5.5 cm

Using Heron’s formula,

Area = \(\sqrt{s(s - a)(s - b)(s - c)}\)

= √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm²

= √[5.5×0.5×0.5×4.5] cm²

= 0.75√11 cm²

= 0.75 × 3.317 cm²

= 2.488 cm² (approx)

For the quadrilateral II section:

This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.

∴ Area = 6.5 × 1 cm² = 6.5 cm²

For the quadrilateral III section:

It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.

Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle

The perpendicular height of the parallelogram will be

= \(\sqrt{1^2 - (0.5)^2}\)

= 0.86 cm

And, the area of the equilateral triangle will be (√3/4 × a²) = 0.43

∴ Area of the trapezoid = 0.86 + 0.43

= 1.3 cm² (approximately).

For triangle IV and V:

These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm

Area triangles IV and V = 2 × (1/2 × 6 × 1.5) cm² 

= 9 cm²

So, the total area of the paper used = (2.488 + 6.5 + 1.3 + 9) cm² 

= 19.3 cm²